f(x) is a polynomial function, `f: R rarr R,` such that `f(2x)=f'(x)f''(x).`
f(x) is

Options

A. 4
B. 12
C. 15
D. none of these

Answer

B

Explanation

Suppose degree of f(x) = n. Then degree of f'=n-1 and degree of `f''=n-2.` So,
`n=n-1+n-2`
Hence, n=3.
So, put `f(x)=ax^(3)+bx^(2)+cx+d (where a ne 0).`
From `f(2x)=f'(x).f''(x),` we have
`8ax^(3)+4bx^(2)+2cx+d=(3ax^(2)+2bx+c)(6ax+2b)`
`=18a^(2)x^(3)+18abx^(2)+(6ac+4b^(2))x+2bc`
Comparing coefficients of terms, we have
`18a^(2)=8a" "rArra=4//9`
`18ab=4b" "rArrb=0`
`2c=6ac+4b^(2)" "rArrc=0`
`d=2bc" "rArrd=0`
`"or "f(x)=(4x^(3))/(9),` which is clearly one-one and onto
`"or "f(3)=12`
`"Also, "(4x^(3))/(9)=xor x=0, x= pm 3//2.`
Hence, sum of roots of equation is zero.